Question

Solve the following systems of equations.
$2x^2+y^2+3xy=12,2(x+y{)}^2-y^2=14$

Answer 1

$2x^2+y^2+3xy=\mathrm{12....}\left(i\right)$

$2(x+y{)}^2-y^2=\mathrm{14...}(ii)$

Simplifying equation (ii)

$2(x^2+y^2+2xy)-y^2=14$

or, $2x^2+2y^2+4xy-y^2=14$

or, $2x^2+y^2+3xy+xy=14$

or, $2x^2+y^2+3xy=14-xy....\left(iv\right)$

Putting the value of $(2x^2+y^2+3xy)$ in equation (i)

$14-xy=12$

or, $14-12=xy$

or, $xy=2$

Hence, every pair of $x$ & $y$ will be the solution of

given equations, which satisfies $xy=2$

So, there will be infinite solutions.